Complex variables conformal mapping trig identity

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SUMMARY

The discussion centers on mapping the function \( w = \left(\frac{z-1}{z+1}\right)^{2} \) for \( z = e^{i\theta} \) where \( \theta \) ranges from 0 to \( \pi \). Participants suggest substituting \( z \) with \( e^{i\theta} \) to simplify the expression, leading to the transformation of the semicircular arc in the complex plane. The hint provided emphasizes the importance of this substitution in reducing the expression to a trigonometric function. The conversation highlights the necessity of multiplying the fraction by \( e^{-i\theta/2} \) to facilitate further simplification.

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  • Familiarity with trigonometric identities and functions
  • Knowledge of exponential forms of complex numbers
  • Basic skills in algebraic manipulation of complex fractions
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EnginerdRuns
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Homework Statement


map the function [tex]\begin{equation}w = \Big(\frac{z-1}{z+1}\Big)^{2} \end{equation}[/tex]
on some domain which contains [tex]z=e^{i\theta}[/tex]. [tex]\theta[/tex] between 0 and [tex]\pi[/tex]
Hint: Map the semicircular arc bounding the top of the disc by putting [tex]$z=e^{i\theta}$[/tex] in the above formula. The resulting expression reduces to a simple trig function.

Homework Equations


I can get the map if I can figure out what function they're going for, but I have no idea what function this is.


The Attempt at a Solution


[tex]$$w = \Big(\frac{e^{i\theta}-1}{e^{i\theta}+1}\Big)^{2}$$[/tex]
Where the heck do I go from here?
 
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Hi EnginerdRuns! Welcome to PF! :smile:

Multiply top and bottom of the fraction by e-iθ/2 :wink:
 
Thanks bro. I'm running off of way too little sleep at this point.
 

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